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Q.
For the reaction, $X_{2} Y_{4}(l) \longrightarrow 2 X Y_{2}(g)$ at $300\, K$, the values of $\Delta U$ and $\Delta S$ are $2\, kcal$ and $20\, cal \,K ^{-1}$ respectively. The value of $\Delta G$ for the reaction is
For the reaction $X_{2} Y_{4}(l) \longrightarrow 2 X Y_{2}(g)$
$\Delta n_{g}=$ number of gaseous products
- number of gaseous reactants
$\Delta n_{g}=2-0=2$
Given that, $\Delta U=2\, kcal$
$\Delta S=20\, cal\, K ^{-1}$
From the formula,
$\Delta H=\Delta U+\Delta n_{g} R T$
Putting the values, we get
$\Delta H =2+\left(\frac{2 \times 2 \times 300}{1000}\right)$
$=3.2\, kcal =3.2 \times 10^{3} cal$
Also, $\Delta G =\Delta H-T \Delta S$
$=3.2 \times 10^{3}-300 \times 20$
$=3.2 \times 10^{3}-6 \times 10^{3}$
$=-2800\, cal$